# Axiom of Choice implies Zorn's Lemma/Proof 2

## Theorem

Let the Axiom of Choice be accepted.

Then Zorn's Lemma holds.

## Proof

Aiming for a contradiction, suppose that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.

By the Axiom of Choice, there is a mapping $f: X \to X$ such that:

- $\forall x \in X: x \prec \map f x$

Let $\CC$ be the set of all chains in $X$.

By the premise, each element of $\CC$ has an upper bound in $X$.

Thus by the Axiom of Choice, there is a mapping $g: \CC \to X$ such that for each $C \in \CC$, $\map g C$ is an upper bound of $C$.

Let $p$ be an arbitrary element of $X$.

Define a mapping $h: \On \to X$ by transfinite recursion thus:

\(\ds \map h 0\) | \(=\) | \(\ds p\) | ||||||||||||

\(\ds \map h {\alpha^+}\) | \(=\) | \(\ds \map f {\map h \alpha}\) | ||||||||||||

\(\ds \map h \lambda\) | \(=\) | \(\ds \map f {\map g {h \sqbrk \lambda} }\) | if $\lambda$ is a limit ordinal |

when $h \sqbrk \lambda$ is the image set of $\lambda$ under $h$.

This article, or a section of it, needs explaining.In particular: If you're defining the mapping $h$, how can you define it in terms of itself? $\map h \lambda = \map f {\map g {h \sqbrk \lambda} }$ looks weird. Might be worth going to the Talk page to discuss.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Then $h$ is strictly increasing, and thus injective.

This article, or a section of it, needs explaining.In particular: The above statement needs to be justified.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Let $h'$ be the restriction of $h$ to $\On \times \map h \On$.

Then ${h'}^{-1}$ is a surjection from $\map h \On \subseteq X$ onto $\On$.

By the Axiom of Replacement, $\On$ is a set.

By the Burali-Forti Paradox, this is a contradiction.

Thus we conclude that some element of $X$ has no strict successor, and is thus maximal.

$\blacksquare$